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$\large \sqrt{5+1\sqrt{6+2\sqrt{ 7+3\sqrt{8+4\sqrt{ 9+5\sqrt{\dots}}}}}} = \ ?$

This problem is a part of set nested radicals .

3 4 2 5 1

1 solution

Ankith A Das
Mar 13, 2015

$\sqrt { 5+1\sqrt { 6+2\sqrt { 7+3\sqrt { 8+4\sqrt { 9+5\sqrt { \dots } } } } } } \\ Using\quad Ramanujan's\quad nested\quad radical\quad formula\quad \\ x+n+a =\sqrt { ax+{ \left( a+n \right) }^{ 2 }+x\sqrt { a\left( x+n \right) +{ \left( a+n \right) }^{ 2 }+\left( x+n \right) \sqrt { ... } } } \\ We\quad can\quad see\quad that\quad x=1\\ then\quad x+n=2\\ 1+n=2\quad ;\quad n=1\\ Substituting\quad x\quad and\quad n\quad in\quad ax+{ \left( a+n \right) }^{ 2 }\\ a\left( 1 \right) +{ \left( a+1 \right) }^{ 2 }=\quad 5\\ a=1\\ x+n+a=1+1+1=3$

We can do this without referring to Ramanujan's formula: Doing some numerical experimentation, we suspect that the answer is 3. We can prove this result by repeatedly using the formula $x=$ $\sqrt{(x+2)+(x-2)(x+1)}$ , always applying it to the last number appearing in an expression: $3=\sqrt{5+1*4}$ = $\sqrt{5+1\sqrt{6+2*5}}$ = $\sqrt{5+1\sqrt{6+2\sqrt{7+3*6}}}$ =...

Otto Bretscher - 6 years, 2 months ago

brilliant!.. the level of this problem should be more than this..

Bhupendra Jangir - 6 years, 3 months ago

true that.

Ankith A Das - 6 years, 3 months ago

It is very easy to approximate the answer as 3, which is why the problem rating is so low.

Daniel Liu - 6 years, 3 months ago

@Daniel Liu – Can you please tell how you approximated the answer.

Ankith A Das - 6 years, 3 months ago

@Ankith A Das – Let's say that $6+2\sqrt{...}\approx 9$ , so $5+\sqrt{6+2\sqrt{...}}\approx 8$ so $\sqrt{5+\sqrt{6+2\sqrt{...}}}\approx 3$

Daniel Liu - 6 years, 3 months ago

Soon to be not-obscure form

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